Calculate the pH of a solution obtained by diluting 25 ml of N/100 HCl to 500 ml.

Explanation

To calculate the pH of a solution obtained by diluting 25 ml of N/100 HCl to 500 ml, we can follow these steps:

### Step 1: Determine the initial normality and volume
We have:
- Initial normality (N1) = N/100 = 0.01 N
- Initial volume (V1) = 25 ml

### Step 2: Determine the final volume after dilution
The final volume (V2) after dilution is given as:
- Final volume (V2) = 500 ml

### Step 3: Use the dilution formula
We can use the dilution formula:
\[ N1 \times V1 = N2 \times V2 \]
Where:
- N2 = final normality after dilution

Substituting the known values:
\[ 0.01 \, \text{N} \times 25 \, \text{ml} = N2 \times 500 \, \text{ml} \]

### Step 4: Solve for N2
Rearranging the equation to solve for N2:
\[ N2 = \frac{0.01 \times 25}{500} \]
\[ N2 = \frac{0.25}{500} \]
\[ N2 = 0.0005 \, \text{N} \]
\[ N2 = 5 \times 10^{-4} \, \text{N} \]

### Step 5: Relate normality to hydrogen ion concentration
Since HCl is a strong acid, the normality (N2) is equal to the concentration of hydrogen ions \([H^+]\):
\[ [H^+] = 5 \times 10^{-4} \, \text{mol/L} \]

### Step 6: Calculate the pH
The pH is calculated using the formula:
\[ \text{pH} = -\log[H^+] \]
Substituting the value of \([H^+]\):
\[ \text{pH} = -\log(5 \times 10^{-4}) \]

### Step 7: Calculate the logarithm
Using logarithmic properties:
\[ \text{pH} = -(\log(5) + \log(10^{-4})) \]
\[ \text{pH} = -(\log(5) - 4) \]
Assuming \(\log(5) \approx 0.699\):
\[ \text{pH} = -(0.699 - 4) \]
\[ \text{pH} = 4 - 0.699 \]
\[ \text{pH} \approx 3.301 \]

### Final Answer
The pH of the diluted solution is approximately **3.3**.

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